Substituting y = x^4, we get
y + √(y+20) = 22
√(y+20) = 22 - y
y + 20 = (22 - y)^2 = 484 - 44y + y^2
y^2 - 45y + 464 = 0
(y - 29)(y - 16) = 0
x^4 = 16 has two integral roots:
X = 2, x = -2
It also has 6 other roots, of which 4 are imaginary, assuming the "under root" can be taken as positive or negative.
y + √(y+20) = 22
√(y+20) = 22 - y
y + 20 = (22 - y)^2 = 484 - 44y + y^2
y^2 - 45y + 464 = 0
(y - 29)(y - 16) = 0
x^4 = 16 has two integral roots:
X = 2, x = -2
It also has 6 other roots, of which 4 are imaginary, assuming the "under root" can be taken as positive or negative.